Question

In $x\ln (\ln x)-{{x}^{2}}+{{y}^{2}}=4(y>0)$ , then $\dfrac{dy}{dx}at\text{ }x=e$ is equal to:
A. $\dfrac{e}{\sqrt{4+{{e}^{2}}}}$
B. $\dfrac{1+2e}{2\sqrt{4+{{e}^{2}}}}$
C. $\dfrac{2e-1}{2\sqrt{4+{{e}^{2}}}}$
D. $\dfrac{1+2e}{\sqrt{4+{{e}^{2}}}}$

Answer 1

Hint: The required expression can be obtained by simply differentiating the given equation; $x\ln (\ln x)-{{x}^{2}}+{{y}^{2}}=4$ with respect to x implicitly i.e. dependent variable appears in the derivative and wise substitution of values at different steps of solution.

Complete step by step answer:
Given equation $x\ln (\ln x)-{{x}^{2}}+{{y}^{2}}=4$ can be differentiated with respect to x implicitly.
$x\ln (\ln x)-{{x}^{2}}+{{y}^{2}}=4\ldots eq(1)$
On rearranging $eq\left( 1 \right)$we get,
\[{{y}^{2}}={{x}^{2}}-x\ln (\ln x)+4\ldots eq(2)\]
 We know differentiate on both side with respect to x. Differentiating above equation need following important formula:
$\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$, $\dfrac{d\left( f\left( x \right).g\left( x \right) \right)}{dx}=f'\left( x \right)g\left( x \right)+f\left( x \right)g'\left( x \right)$, $\dfrac{d(f\left( g\left( x \right) \right)}{dx}={f}'\left( g\left( x \right) \right){g}'\left( x \right)$, $\dfrac{d\left( x\ln \left( \ln \left( x \right) \right) \right)}{dx}=\ln \left( \ln \left( x \right) \right)+\dfrac{1}{\ln \left( x \right)}$, $\dfrac{d\left( c \right)}{dx}=0$
$\dfrac{d}{dx}({{y}^{2}})=\dfrac{d}{dx}({{x}^{2}}-x\ln (\ln (x))+4)$
$\dfrac{d}{dx}\left( {{y}^{2}} \right)=\dfrac{d}{dx}\left( {{x}^{2}} \right)-\dfrac{d}{dx}\left[ x\ln (\ln (x)) \right]+\dfrac{d}{dx}\left( 4 \right)$
$2y\dfrac{dy}{dx}=2x-\ln (\ln x)-\dfrac{1}{\ln x}+0$
$\dfrac{dy}{dx}=\dfrac{1}{2y}\left( 2x-\ln (\ln x)-\dfrac{1}{\ln x} \right)\ldots eq(3)$
We can obtain the value of y from the original equation $x\ln (\ln x)-{{x}^{2}}+{{y}^{2}}=4$by substituting the value of $x=e$,
$eln(ln(e))-{{e}^{2}}+{{y}^{2}}=4$
As we know that$\ln (\ln (e))=1,\ln (1)=0$ we get following value of y,
$y=\sqrt{{{e}^{2}}+4}$
At$x=e$, the $eq\left( 3 \right)$gives,
$\dfrac{dy}{dx}=\dfrac{1}{2y}\left( 2e-1 \right)$
$\dfrac{dy}{dx}=\dfrac{2e-1}{2\sqrt{{{e}^{2}}+4}}at\text{ }x=e$

So, the correct answer is “Option C”.

Note: The implicit differentiation must be done properly. The equation $x\ln (\ln x)-{{x}^{2}}+{{y}^{2}}=4$can be implicitly differentiated without rearranging the terms (rearranging is just for simplicity) as given below:
$\dfrac{d}{dx}(x\ln (\ln x)-{{x}^{2}}+{{y}^{2}})=\dfrac{d}{dx}(4)$
$\dfrac{1}{\ln x}+\ln (\ln (x))-2x+2y\dfrac{dy}{dx}=0$